(a) Derive the expression for the torque acting on the rectangular current carrying coil of a galvanometer. Why is the magnetic field made radial?
(b) An $\mathrm{\alpha }$-particle is accelerated through a potential difference of 10 kV and moves along x-axis.
It enters in a region of uniform magnetic field B = 2 x 10^–3 T acting along y-axis. Find the radius of its path. (Take mass of $\mathrm{\alpha }$-particle = 6·4 x 10^–27 kg )

(a) With the help of a labelled diagram, explain the working of a step-up transformer. Give reasons to explain the following :
(i) The core of the transformer is laminated.
(ii) Thick copper wire is used in windings.

Magnetic forces of AB and CD are equal and opposite and have different line of action so constitute
torque
Force acting on current carrying arms AB and CD
$F_1 = F_2 = 𝐵𝐼𝑙 = 𝐹 (𝑠𝑎𝑦)$
$\therefore \mathrm{\tau }=\mathrm{F}\times$𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑙𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑓𝑜𝑟𝑐𝑒 𝑎𝑟𝑚
$\therefore \mathrm{\tau }=\mathrm{BIlbsin} \mathrm{\theta lb}=\mathrm{A\tau }=\mathrm{BIAsin} \mathrm{\theta }$
For N turn $\mathrm{\tau }=\mathrm{BINAsin} \mathrm{\theta }$
Radial fields always produce maximum torque and removes the dependence of torque on $\mathrm{\theta }$
(a) Radius of circular path$=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{\sqrt{2{\mathrm{mE}}_{\mathrm{k}}}}{\mathrm{Bq}}$
$\begin{array}{l}=\frac{1}{\mathrm{B}}\frac{\sqrt{2\mathrm{mq}}}{{\mathrm{q}}^2}=\frac{1}{\mathrm{B}}\frac{\sqrt{2\mathrm{mV}}}{{\mathrm{q}}^2}=\frac{1}{2\ast {10}^{-3}}\\ \mathbf{r}=10\mathrm{m}\end{array}$
OR
When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf Induced emf across primary coil
${\mathrm{e}}_{\mathrm{P}}=-{\mathrm{N}}_{\mathrm{p}}\frac{\mathrm{d}\mathrm{\varnothing }}{\mathrm{dt}}$
Induced emf across secondary coil
${\mathrm{e}}_{\mathrm{s}}=-{\mathrm{N}}_{\mathrm{s}}\frac{\mathrm{d}\mathrm{\varnothing }}{\mathrm{dt}}$
$\frac{{\mathrm{e}}_{\mathrm{S}}}{{\mathrm{e}}_{\mathrm{p}}}=\frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}}=\mathrm{r}$
(i) To minimize the eddy currents (ii) To reduce the heat loss
(b) A conducting rod PQ of length 20 cm and resistance 0·1 $\mathrm{\Omega }$ rests on two smooth parallel rails of negligible resistance AA and CC. It can slide on the rails and the arrangement is positioned between the poles of a permanent magnet producing uniform magnetic field B = 0·4 T. The rails, the rod and the magnetic field are in three mutually perpendicular directions as shown in the figure. If the ends A and C of the rails are short circuited, find the
(i) external force required to move the rod with uniform velocity v = 10 cm/s, and
(ii) power required to do so.

$\left(\mathrm{i}\right) \mathrm{F}=\mathrm{BI}l$
$\begin{array}{l}\mathrm{I}=\mathrm{E}/\mathrm{R}=\mathrm{Bv}l/\mathrm{R}\\ \mathrm{F}={\mathrm{B}}^2{\mathrm{vl}}^2/\mathrm{R}\\ =\frac{0.\ddot{4}\times 0.4\times 0.1\times 0.2\times 0.2}{0.1}\\ =6.4\times {10}^{-3}\mathrm{N}\\ \begin{array}{c}\mathrm{P}=\mathrm{F}.\mathrm{v}=6.4\times {10}^{-3}\times 0.1\\ =.64\times {10}^{-3}\mathrm{W}\end{array}\end{array}$